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Q.
If quadratic equation $x^2+2(a+2 b) x+(2 a+b-1)=0$ has unequal real roots for all $b \in R$ then the possible values of a can be equal to
Complex Numbers and Quadratic Equations
Solution:
The quadratic equation $x ^2+2( a +2 b )+(2 a + b -1)=0 $ will have unequal real roots, if $D=4(a+2 b)^2-4(2 a+b-1)>0$
$\Rightarrow a^2+4 b^2+4 a b>2 a+b-1 $
$\Rightarrow 4 b^2+(4 a-1) b+\left(a^2-2 a+1\right)>0 \forall b \in R$
$\therefore (4 a-1)^2-16\left(a^2-2 a+1\right)<0$
$\Rightarrow \left(16 a^2-8 a+1\right)-\left(16 a^2-32 a+16\right)<0$
$\Rightarrow 24 a-15<0$
Hence $a< \frac{15}{24}$.
Now verify alternatives.